Let $R$ and $S$ be rings, $M$ be an $R$-module, $N$ be an $S$-module, and $\varphi:R\sqcup M\rightarrow S\sqcup N$ be a map satisfying:

  1. $\varphi\restriction_R:R\rightarrow S$ is a ring homomorphism
  2. $\varphi\restriction_M:M\rightarrow N$ is a group homomorphism
  3. $\varphi(am)=\varphi(a)\varphi(m)$ for all $a\in R, m\in M$

In the special case where $R=S$ and $\varphi\restriction_R$ is the identity map on R, we call $\varphi\restriction_M$ an $R$-module homomorphism. Is there something I am missing that makes $R$-module homomorphisms the only interesting example of such maps, or is this a worthy generalization? Could we define the category Mod of modules using these maps as the morphisms?

  • $\begingroup$ What a strange notation. So just to clarify, what you really have is a map $\varphi_1 : R \to S$ and a map $\varphi_2 : M \to N$...? Why would you denote such a thing like that? $\endgroup$ – Najib Idrissi Nov 21 '15 at 12:24
  • $\begingroup$ It's essentially two maps, but combining them as I did makes property 3. easier to express and so it just made sense to me. $\endgroup$ – ngenisis Nov 21 '15 at 17:45
  • $\begingroup$ Equation 3 would simply have been "$\varphi_2(am) = \varphi_1(a) \varphi_2(m)$"... Not very complicated, and instead you have to remember that $\varphi$ is really two different maps. $\endgroup$ – Najib Idrissi Nov 21 '15 at 17:54
  • $\begingroup$ To each his own... I think oxeimon had the best description with the pullback module $\endgroup$ – ngenisis Nov 21 '15 at 19:13
  • $\begingroup$ Sure, to each his own. Just be aware that your notation is very nonstandard and most people will be confused by it. oxeimon's notation, OTOH, is standard. $\endgroup$ – Najib Idrissi Nov 21 '15 at 19:19

What you have described is basically the "pullback" of $N$ to the category of $R$-modules via the homomorphism $\varphi|_R : R\rightarrow S$.

Also, I don't like your $\sqcup$ notation. Really you're specifying the data of two things:

  1. A homomorphism $\varphi : R\rightarrow S$, and
  2. An $R$-linear map $M\rightarrow \varphi^*N$, where $\varphi^*N$ is just $N$ viewed as an $R$-module via $\varphi$ (ie, for $r\in R$, set $r.n = \varphi(r).n$)

What you're describing is precisely the category of affine schemes equipped with a quasicoherent sheaf of modules. I don't think this is particularly useful unless you specify some additional conditions on which $R,M,S,N$ are allowed.


What is interesting is the pair consisting of this category and its forgetful functor to rings; this exhibits the functor $R \mapsto \text{Mod}(R)$ as a fibered category or Grothendieck fibration.

  • $\begingroup$ Oh I guess that's just the moduli stack of quasicoherent sheaves? $\endgroup$ – oxeimon Nov 21 '15 at 7:30

Have you ever heard of restriction of scalars? It works as follows: if $f:R\longrightarrow S$ is a homomorphism of rings then any left (right) $S$-module $N$ inherits a left (right) $R$-module structure if we define $$r\cdot x:=f(r)\cdot x\quad (x\in N, r\in R).$$ Now, given a left $R$-module $M$ and a left $S$-module $N$ then we can define a homomorphism over $f$ as a homomorphism of $R$-modules $u: M\longrightarrow N$ (where the $N$ is endowed with the $R$-module structure induced by $f$ as described above). So, in my opinion, there is no point in introducing that notion of homomorphism because it can be reduced to the usual one using modules over the same ring.


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